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Math 3704 Lecture Notes Math 3704 Lecture Notes Minhyong Kim, based on notes by Richard Hill October 13, 2008 Please let me ([email protected]) know of any misprints or mistakes that you ¯nd. Contents 1 Introduction 2 1.1 Prerequisites for the course . 3 1.2 Course Books . 3 2 Background Material 3 2.1 Polynomial Rings . 3 2.1.1 Euclid's Algorithm . 7 2.1.2 Ideals . 7 2.1.3 Quotient rings . 8 2.1.4 Homomorphisms of rings . 8 2.2 Field extensions . 9 2.3 Degrees of extensions . 11 2.4 Symmetric polynomials . 14 2.5 k-Homomorphisms * . 15 2.6 Splitting ¯elds and Galois groups * . 16 2.7 Calculating Galois groups * . 17 3 Algebraic Number Fields 18 3.1 Field embeddings . 18 3.2 Norm, Trace and Discriminant . 20 3.3 Algebraic Integers . 23 3.4 Integral Bases . 25 3.5 Integral bases in quadratic ¯elds . 27 3.6 Cubic ¯elds . 28 3.7 More tricks for calculating integral bases . 32 3.8 More examples of integral bases . 33 3.9 Prime Cyclotomic Fields . 34 4 Factorization in ok 36 4.1 Units and irreducible elements in ok .............................. 36 4.2 Prime ideals . 39 4.3 Uniqueness of Factorization into ideals . 41 4.4 Norms of ideals . 46 4.5 Norms of prime ideals . 48 4.6 Factorizing Ideals into Maximal Ideals . 51 4.7 The Class Group . 51 4.8 The Minkowski constant . 52 1 4.9 Geometry of numbers and Minkowski's Lemma . 53 4.10 The Minkowski Space . 54 4.11 Calculating class groups . 57 4.12 Dirichlet's Unit Theorem . 60 2 Lecture 1 1 Introduction Three ¯elds that occur in nature are Q, the ¯eld of rational numbers, the ¯eld R of real numbers, and the ¯eld C of complex numbers. The ¯rst ¯eld is eventually forced on you as you proceed through arithmetic operations on counting numbers. The latter two arise in describing rigorously the continuous objects of the universe, as well as the microscopic world. In spite of many outstanding problems in number theory, the internal structure of Q is in some sense well-motivated and clear. However, the construction of R and C are mysterious. These are extremely large objects comprising many layers of complexity that are constructed, in some sense, all at one go starting from the rationals. Eventually, there arises a need to bridge the enormous gap visible in the inclusion Q ½ C: One way to think about this issue is in terms of various intermediate objects Q ½ F1 ½ F2 ½ ¢ ¢ ¢ ½ C that we try to construct at a slower pace, keep track of various properties as we go. Many such interme- diate ¯elds are of great interest, but a good starting point is to consider intermediate ¯elds consisting of algebraic numbers. Algebraic numbers are special or even universal in that copies of these numbers exist inside any su±ciently rich number system wherein we can count naturally. A number ® 2 C is said to be algebraic if there is a non-zero polynomial f 2 Q[X] such that f(®) = 0. An algebraic number ¯eld is a ¯eld of the form ½ ¾ f(®) Q(®) = : f; g 2 Q[X]; g(®) 6= 0 ; g(®) i.e. the ¯eld generated by Q and ®. For example p p Q( 2) = fx + y 2 : x; y 2 Qg; p p p Q( 3 2) = fx + y 3 2 + z 3 4 : x; y 2 Qg: In any algebraic number ¯eld k there is a ring of algebraic integers o. An algebraic number is called an algebraic integer, if it is a zero of a monic polynomial with integer coe±cients. Examples of these rings of algebraic integers are: p p Z[ 2] ½ Q( 2); where Z[®] = ff(®): f 2 Z[X]g: The sort of questions that we'll deal with in this course are: ² Does the ring o have unique factorization? ² Is o a principal ideal domain? ² If o is not a principal ideal domain, then how far is it from being a principal ideal domain? ² If p is a prime number, how does p factorize in o? For example in Z[i] we have 5 = (2 + i)(2 ¡ i), but 7 does not factorize in Z[i]. p p p p ² What are the units in o? For example inp Z[ 2] we have ( 2 + 1)( 2 ¡ 1) = 1, so 2 + 1 is a unit. On the other hand the only units in Z[ ¡5] are 1 and ¡1. 3 Instead of factorizing elements of o we shall factorize ideals. If o is a principal ideal domain then this is the same thing. We shall show that every ideal of o can be uniquely factorized into prime ideals of o. Therefore if o is a principal ideal domain then we have uniqueness of factorization of elements. If o is not a principal ideal domain then we can measure how far it is from being a principal ideal domain by calculating the class group: Cl = fidealsg=fprincipal idealsg: This turns out to be a ¯nite group that measures the complexity of the ¯eld k and the ring o in pretty much the same way that the homology groups in algebraic topology measure the complexity of a space. A rather speci¯c aim of the course is for you to be able to calculate this group for some simple algebraic number ¯elds. Eventually, you should try to produce yourself ¯elds of various complexity. 1.1 Prerequisites for the course Elementary linear algebra. Group theory and ring theory from MATH 7202, in particular ideals, quotient rings, polynomial rings over a ¯eld. A willingness to think flexibly with diverse mathematical notions. 1.2 Course Books There are many books on algebraic number theory. The one by Stewart and Tall listed on the syllabus is a rather elementary introduction intended to be user-friendly at the undergraduate level. There is a book by Ireland and Rosen `A classical invitation to modern number theory' that attempts to reach a similar readership, but contains a more sophisticated viewpoint. A book called `Fermat's Dream' by Kazuya Kato et. al. adopts a highly inspirational approach emphasizing the role of zeta functions. `A course in arithmetic' by J.-P. Serre deals with some rather sophisticated topics in a self-contained way. There is a set of online notes by James Milne available at www.jamesmilne.org that develops algebraic number theory in a fairly systematic manner at the post-graduate level. 2 Background Material 2.1 Polynomial Rings In this course all rings will be commutative rings with 1. Let k be a ¯eld. We shall write k[X] for the ring of polynomials in the variable X with coe±cients in k. Recall that in any ring R there are three kinds of elements: ² The units; ² The reducible elements; ² The irreducible elements. 1 Proposition k[X] is an integral domain Proof. Let f(X) and g(X) be non-zero. Then they have the form n f(X) = a0 + a1X + ¢ ¢ ¢ + anX and m g(X) = b0 + b1X + ¢ ¢ ¢ + bmX with an 6= 0 and bm 6= 0. But then the highest term of f(X)g(X) is anbm 6= 0. So f(X)g(X) 6= 0. 2 4 We will frequently take a polynomial f(X) 2 Z[X] and consider its reduction modulo p for a prime p. ¹ This is the polynomial f(X) 2 Fp[X] obtained by reducing all the coe±cients of f mod p. For example, if f(X) = 5X4 + 9X3 + 2X + 3, then for the prime 3, we have ¹ 4 f = [2]X + [2]X 2 F3[X]: Here, we are writing [2] for the congruence class in F3 of 2. But much of the time, we will omit the square brackets when the context makes the reduction clear. Also, in the notation f¹, we will not indicate the prime p separately. It also will be indicated by the context. For any polynomial f(X) 2 Q[X], we will denote by f1(X) 2 Z[X] the unique constant multiple of f with the property that (1) f1 is primitive, that is, its coe±cients are coprime. (2) Its leading coe±cient is positive. One obtains f1(X) from f(X) by ¯rst multiplying f by an integer c so that cf(X) 2 Z[X]. One then divides by the highest common factor r of the coe±cients of cf to obtain (c=r)f(X) 2 Z[X] primitive. One then multiplies (c=r)f by 1 or ¡1 in order to make its leading coe±cient positive. For example, if f(X) = (¡4)X2 + (2=3)X + 10; then one goes to (¡12)X2 + 2X + 30 to ¡6X2 + X + 15 to 2 f1 = 6X ¡ X ¡ 15: Exercise: Given f 2 Q[X] show that f1(X) with the two properties above is unique. In the ring k[X] the units are the non-zero constant polynomials, i.e. the elements of k£. There are various ways of deciding whether elements of Q[X] are irreducible or not. 2 Gauss' Lemma Suppose f 2 Z[X] and assume that f is not constant. If f is irreducible as an element of Z[X] then f is irreducible as an element of Q[X]. Proof. Suppose f(X) = g(X)h(X) in Q[X]. We have g = ag1 and h = bh1 for some constants a; b 2 Q, so that f(X) = abg1(X)h1(X): Claim: g1(X)h1(X) is primitive. Suppose some p divided all the coe±cients of g1(X)h1(X).
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